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Internal energy enthalpy problem

  1. Jan 8, 2010 #1
    hello happy new year

    i'm a little confused here after i saw this

    U= KE + PE + H(enthalpy)----------------------1

    while it is known that

    i'm sure that eqn 2 is true but do you think with m that 1 is not true ??

    the problem is that my instructor use the eqn 1 in 1st law

    Q-W= delta(KE +PE+H)
  2. jcsd
  3. Jan 8, 2010 #2


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    Hi aeroguy. Your instructor is correct. Kinetic and potential energy can be added to the first law. Often these are neglected because they're small, but when the fluid has a very high velocity, kinetic energy can be significant and shouldn't be neglected.
  4. Jan 8, 2010 #3
    Though it is correct, as Q_Goest said, I think it is a poor choice of symbols. It would be better written using something like E for energy:

    [tex]\Delta E = Q - W = \Delta KE + \Delta PE + \Delta U[/tex]

    for a closed system
  5. Jan 9, 2010 #4
    hey guys
    i guess you misunderstand me

    i ask about the U:
    U =delta KE + delta PE + ENTHALPY

    my problem is it true that
    U= (KE+PE)+ ENTHALPY <===========

  6. Jan 9, 2010 #5


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    Hi aero... you're right. I misunderstood you. Kinetic and potential energy are not part of the overall internal energy, they are in addition to it.

    Take a look at this web page:

    Scroll down just a little (about 10% of the way down) to paragraph I.A.3. There's a blue circle with a question mark (?). Just above where that question mark is, they show the expanded relationship for the first law which includes kinetic, potential and also chemical energy. Kinetic and potential energy are not part of internal energy, they are in addition to it.

    Note that Yeti also shows the correct relationship above.

    The OP, equation 1 is incorrect:
    U= KE + PE + H(enthalpy)----------------------1
    Last edited by a moderator: Apr 24, 2017
  7. Jan 9, 2010 #6
    No, I understood your question which is why I pointed out that the correct equation should use the change in energy, not internal energy. My point about symbols is that you could define U to mean anything you want, but you need to be consistent; your example equations are not consistent. Additionally, you're using enthalpy in your equation which is only used when flow work is involved (i.e. a control volume with mass flowing past its boundaries) which for a given control volume with inlets (i) and exits (e) is, in general form:

    [tex]\frac{dE}{dt}= \dot{Q}-\dot{W}+

    \sum_{i}{\dot{m_{i}}\left(\ h_{i}+\frac{vel_{i}^{2}}{2}+g\:z_{i}\right)}}-\sum_{e}{\dot{m_{e}}\left(\ h_{e}+\frac{vel_{e}^{2}}{2}+g\:z_{e}\right)}}[/tex]

    You can see the enthalpy, KE, and PE terms in each stream in the equation. The equation I gave in my last post was for a closed system.
    Last edited: Jan 10, 2010
  8. Jan 9, 2010 #7


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    Hi Yeti. Thanks for that, it's an excellent post, and very nicely explained. Just a thought and a nitpick. I think it would help to also explain what you mean by open and closed system. I don't disagree with you but I think many students have a problem distinguishing this. Also, you need to add a negative sign before the second summation symbol. If I knew Latex, I'd write it out, but I don't. :blushing:
  9. Jan 10, 2010 #8
    Of course I would leave out an important math operation...

    As far as an open versus a closed system in thermodynamics (if anyone is interested). The simplest way to think of it is a closed system will allow heat and work to pass through the boundary, but not mass flow. An open system will allow all - work, heat, and mass - to pass through, but the boundary, called a control volume, must be more carefully chosen with the needed calculations in mind. An example of a closed system could be a piston and sealed cylinder - note that the piston represents a moving boundary, which is allowed. An open system example would be a water heater with incoming cold water, exiting hot water, and added heat.
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